Finding the mean of normal distribution through integration over $[a, b]$

Question

I know the formula to finding this mean is to integrate $x\frac{1}{b}$ from $a$ to $b$. Can someone explain why this is so? I've been trying to compute the mean with the standard formula ($\int_a^b xf(x)dx$) but it's been horribly painful. Thanks! The bound of the distribution is $[0, b]$

Answer 1

I thtink you mean, that X is uniformly distributed: $X\sim U(a,b)$. In your case $X\sim U(0,b)$

$f(x)=\begin{cases} \frac{1}{b}, \ 0\leq x \leq b \\ 0, \ \text{elsewhere}\end{cases}$

In general the mean of a continuous random variable is $E(x)=\int_{-\infty}^{\infty} x\cdot f(x) \ dx$

Thus the expected value of X is

$E(X)=\int_0^b x \frac{1}{b} \ dx$

$\frac{1}{b}$ is a constant. Therefore it you can be factored out.

$E(x)=\frac{1}{b} \int_0^b x \ dx$

I think you can proceed from here.